已知$S(n,k)=\displaystyle \sum_{i=1}^n{i^k}$，其中$k,n\in\mathbb N^*$．
(1)求$S(n,1)$，$S(n,2)$，$S(n,3)$；
(2)给出$S(n,k)$关于$k$的一个递推公式．
分析与解 (1) 根据等差数列的求和公式，有\[S(n,1)=\dfrac 12n(n+1)=\dfrac 12n^2+\dfrac 12n.\]由和立方公式，有\[(i+1)^3=i^3+3i^2+3i+1,\]于是\[\sum_{i=1}^n(i+1)^3=\sum_{i=1}^ni^3+3S(n,2)+3S(n,1)+n,\]即\[(n+1)^3-n-1=3S(n,2)+3S(n,1),\]整理可得\[S(n,2)=\dfrac 13n^3+\dfrac 12n^2+\dfrac 16n.\]类似的，由\[(i+1)^4=i^4+4i^3+6i^2+4i+1,\]可得\[S(n,3)=\dfrac 14n^4+\dfrac 12n^3+\dfrac 14n^2.\]
(2) 由二项式定理，有\[(i+1)^{k+1}=i^{k+1}+{\rm C}_{k+1}^ki^k+{\rm C}_{k+1}^{k-1}i^{k-1}+\cdots+{\rm C}_{k+1}^1i+1,\]于是\[\sum_{i=1}^n(i+1)^{k+1}=\sum_{i=1}^ni^{k+1}+{\rm C}_{k+1}^kS(n,k)+{\rm C}_{k+1}^{k-1}S(n,k-1)+\cdots+{\rm C}_{k+1}^1S(n,1)+n,\]整理可得\[S(n,k)=\dfrac{(n+1)^{k+1}-n-1-{\rm C}_{k+1}^{k-1}S(n,k-1)-{\rm C}_{k+1}^{k-2}S(n,k-2)-\cdots-{\rm C}_{k+1}^1S(n,1)}{k+1}.\]
事实上，有\[S(n,k)=\dfrac{1}{k+1}\sum_{i=1}^{k+1}{\rm C}_{k+1}^iB_{k+1-i}n^i,\]其中\[\dfrac{z}{{\rm e}^z-1}=\sum_{n=0}^{+\infty}B_n\dfrac{z^n}{n!}.\]中间用到的$B_n$被称为伯努利数(Bernoulli number).
观察下列等式：
\begin{align*}
\displaystyle\sum_{i=1}^{n}i&=\dfrac{1}{2}n^2+\dfrac{1}{2}n,\\
\displaystyle\sum_{i=1}^{n}i^2&=\dfrac{1}{3}n^3+\dfrac{1}{2}n^2+\dfrac{1}{6}n,\\
\displaystyle\sum_{i=1}^{n}i^3&=\dfrac{1}{4}n^4+\dfrac{1}{2}n^3+\dfrac{1}{4}n^2,\\
\displaystyle\sum_{i=1}^{n}i^4&=\dfrac{1}{5}n^5+\dfrac{1}{2}n^4+\dfrac{1}{3}n^3-\dfrac{1}{30}n,\\
\displaystyle\sum_{i=1}^{n}i^5&=\dfrac{1}{6}n^6+\dfrac{1}{2}n^5+\dfrac{5}{12}n^4-\dfrac{1}{12}n^2,\\
\displaystyle\sum_{i=1}^{n}i^6&=\dfrac{1}{7}n^7+\dfrac{1}{2}n^6+\dfrac{1}{2}n^5-\dfrac{1}{6}n^3+\dfrac{1}{42}n,\\
&\vdots\\
\displaystyle\sum_{i=1}^{n}i^k&=a_{k+1}n^{k+1}+a_kn^k+a_{k-1}n^{k-1}+a_{k-2}n^{k-2}+\cdots+a_1n+a_0,
\end{align*}可以推测，当$k \geqslant 2\ \left(k\in\mathbb{N}^{*}\right)$时，$a_{k+1}=\dfrac{1}{k+1}$，$a_k=\dfrac{1}{2}$，
$a_{k-1}=$______，$a_{k-2}=$______．
这道填空题考查合情推理．观察出答案很容易：$a_{k-1}=\dfrac{k}{12}$，$a_{k-2}=0$．下面用数学归纳法给出证明．
为表述方便，设
\[
S_k(n)=\displaystyle\sum_{i=1}^{n}i^k=a_{k+1}^{(k)}n^{k+1}+a_k^{(k)}n^k+a_{k-1}^{(k)}n^{k-1}+a_{k-2}^{(k)}n^{k-2}+\cdots+a_1^{(k)}n+a_0^{(k)},
\]
我们现在要证明对任意$k \geqslant 2\ \left(k\in\mathbb{N}^{*}\right)$，都有\[\begin{cases}
a_{k+1}^{(k)}=\dfrac{1}{k+1},\\
a_k^{(k)}=\dfrac{1}{2},\\
a_{k-1}^{(k)}=\dfrac{k}{12},\\
a_{k-2}^{(k)}=0.
\end{cases}\]
由题意，当$k=2,3,4$时，欲证等式均成立．
假设欲证等式对$k=m-1,m,m+1$均成立，则有
\begin{align*}
S_{m+1}(n)&=\dfrac{1}{m+2}n^{m+2}+\dfrac{1}{2}n^{m+1}+\dfrac{m+1}{12}n^{m}+\cdots+a_0^{(m+1)},\\
S_{m}(n)&=\dfrac{1}{m+1}n^{m+1}+\dfrac{1}{2}n^{m}+\cdots+a_0^{(m)},\\
S_{m-1}(n)&=\dfrac{1}{m}n^m+\cdots+a_0^{(m-1)},
\end{align*}
因此$k=m+2$时，由递推公式可知
\begin{align*}
a_{m+3}^{(m+2)}&=\dfrac{1}{m+3},\\
a_{m+2}^{(m+2)}&=\dfrac{1}{m+3}\left(\mathrm{C}_{m+3}^{1}-\dfrac{\mathrm{C}_{m+3}^{2}}{m+2}\right)=\dfrac{1}{2},\\
a_{m+1}^{(m+2)}&=\dfrac{1}{m+3}\left(\mathrm{C}_{m+3}^{2}-\dfrac{\mathrm{C}_{m+3}^{2}}{2}-\dfrac{\mathrm{C}_{m+3}^{3}}{m+1}\right)=\dfrac{m+2}{12},\\
a_{m+1}^{(m+2)}&=\dfrac{1}{m+3}\left(\mathrm{C}_{m+3}^{3}-\dfrac{(m+1)\mathrm{C}_{m+3}^{2}}{12}-\dfrac{\mathrm{C}_{m+3}^{3}}{2}-\dfrac{\mathrm{C}_{m+3}^{4}}{m}\right)=0,
\end{align*}
故$k=m+2$时，欲证等式也成立．
综上所述，欲证等式对任意$k \geqslant 2\ \left(k\in\mathbb{N}^{*}\right)$都成立．